The number of distinct real values of `lamda` for which `(x-1)/1=(y-2)/2=(z+3)/(lamda^(2))` and `(x-3)/1=(y-2)/(lamda^(2))=(z-1)/2` are coplanar, is

Options

A. 2
B. 4
C. 3
D. 1

Answer

To determine the number of distinct real values of \(\lambda\) for which the given lines are coplanar, we will follow these steps:

### Step 1: Identify the equations of the lines
We have two lines given in the symmetric form:
1. Line 1: \(\frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{\lambda^2}\)
2. Line 2: \(\frac{x-3}{1} = \frac{y-2}{\lambda^2} = \frac{z-1}{2}\)

### Step 2: Extract points and direction ratios
From the equations, we can extract the points and direction ratios for both lines.

For Line 1:
- Point \(P_1(1, 2, -3)\)
- Direction ratios: \(a_1 = 1\), \(b_1 = 2\), \(c_1 = \lambda^2\)

For Line 2:
- Point \(P_2(3, 2, 1)\)
- Direction ratios: \(a_2 = 1\), \(b_2 = \lambda^2\), \(c_2 = 2\)

### Step 3: Use the coplanarity condition
The lines are coplanar if the following determinant is equal to zero:

\[ \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0 \]

Substituting the values:
- \(x_2 - x_1 = 3 - 1 = 2\)
- \(y_2 - y_1 = 2 - 2 = 0\)
- \(z_2 - z_1 = 1 - (-3) = 4\)

The determinant becomes:

\[ \begin{vmatrix} 2 & 0 & 4 \\ 1 & 2 & \lambda^2 \\ 1 & \lambda^2 & 2 \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Calculating the determinant:

\[ = 2 \begin{vmatrix} 2 & \lambda^2 \\ \lambda^2 & 2 \end{vmatrix} - 0 + 4 \begin{vmatrix} 1 & 2 \\ 1 & \lambda^2 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} 2 & \lambda^2 \\ \lambda^2 & 2 \end{vmatrix} = 2 \cdot 2 - \lambda^2 \cdot \lambda^2 = 4 - \lambda^4\)

2. \(\begin{vmatrix} 1 & 2 \\ 1 & \lambda^2 \end{vmatrix} = 1 \cdot \lambda^2 - 1 \cdot 2 = \lambda^2 - 2\)

Substituting back into the determinant:

\[ = 2(4 - \lambda^4) + 4(\lambda^2 - 2) = 8 - 2\lambda^4 + 4\lambda^2 - 8 = -2\lambda^4 + 4\lambda^2 \]

Setting the determinant to zero for coplanarity:

\[ -2\lambda^4 + 4\lambda^2 = 0 \]

### Step 5: Factor the equation
Factoring out \(-2\lambda^2\):

\[ -2\lambda^2(\lambda^2 - 2) = 0 \]

This gives us:

1. \(-2\lambda^2 = 0 \Rightarrow \lambda^2 = 0 \Rightarrow \lambda = 0\)
2. \(\lambda^2 - 2 = 0 \Rightarrow \lambda^2 = 2 \Rightarrow \lambda = \pm \sqrt{2}\)

### Step 6: Count distinct values
The distinct real values of \(\lambda\) are:
- \(\lambda = 0\)
- \(\lambda = \sqrt{2}\)
- \(\lambda = -\sqrt{2}\)

Thus, there are **three distinct real values** of \(\lambda\).

### Final Answer
The number of distinct real values of \(\lambda\) is **3**.

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